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Chapter 4: Q47E (page 137)

An electron in an atom can "jump down" from a higher energy level to a lower one, then to a lower one still. The energy the atom thus loses at each jump goes to a photon. Typically, an electron might occupy a level for a nanosecond. What uncertainty in the electron's energy does this imply?

Short Answer

Expert verified

The uncertainty in the electron’s energy is ∆E≥5.3×10-23J.

Step by step solution

01

Given data

Time is given as: ∆t=10-9s.

02

Uncertainty principle

The principle states that the position and the velocity of an object cannot be measured with 100 % accuracy at the same time.

∆x·∆p≥h2

∆x= Uncertainty in the position.

∆p = Uncertainty of momentum.

h = Planck's constant. = 1.05x10-34 J.s

03

Electron’s energy

The Energy and Time Uncertainty Principle,

∆t·∆E≥h2

Substituting values, and we get:

role="math" localid="1658386938701" ∆E≥1.05×10-342×1×10-9∆E≥5.3×10-26J

Therefore, the total energy is ∆E≥5.3×10-26J.

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Most popular questions from this chapter

Electrons are accelerated through a 20 V potential difference producing a monoenergetic beam. This is directed at a double-slit apparatus of 0.010 mm slit separation. A bank of electron detectors is 10 m beyond the double slit. With slit 1 alone open, 100 electrons per second are detected at all detectors. With slit 2 alone open, 900 electrons per second are detected at all detectors. Now both slits are open.

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